Lesson 5.5 Homework Solutions
1.
The resultant amplitude is given by:
2 cos
2 0.02 cos
2
12
0.039 mm
s
A
δ
π
=
= ×
×
=
2.
(i)
Resultant amplitude =
2 cos
2 0.05 cos
2
4
0.07 mm
A
= ×
×
=
(ii)
Resultant amplitude =
2 cos
2 0.05
0
cos
2
2
A
= ×
×
=
3.
(i)
1
340 ms
68
z
.
H
5 0
v
f
m
λ
=
=
=
(ii)
Path difference
∆
x = 11.2 – 10 = 1.2 m
1.2 m
5 m
0.24
x
∆
=
=
(iii)
The phase difference that corresponds to a path difference of
λ
= 2
π
Phase difference for 0.24
λ
=
0.48
π
(iv)
The resultant amplitude =
2 cos
2 0.01 cos0.2
0.015
4
2
m
A
m
= ×
×
=
4.
40.07
40
2.4
Path difference 40.07 – 40
= 0.07 m
(i)
The condition for minimum sound is:
(2
1)
0.07
2
n
+
=
For this condition, the frequency will be minimum when n = 0
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View Full DocumentThen,
λ
/2 = 0.07,
or
λ
= 0.14 m
340
2429
0.14
v
f
Hz
λ
=
=
=
The condition for maximum sound is n
λ
= 0.07
The minimum frequency for this condition occurs when n = 1
Then
λ
= 0.07
340
0.07
4857
v
Hz
f
=
=
=
5.
442 Hz or 438 Hz
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 Fall '10
 George
 Physics, Work, 0.07 m, 0.005 m, δ δ, 0.07 mm

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